3.141 592 653 689 802 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 689 802 5(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
3.141 592 653 689 802 5(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 3.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.141 592 653 689 802 5.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.141 592 653 689 802 5 × 2 = 0 + 0.283 185 307 379 605;
  • 2) 0.283 185 307 379 605 × 2 = 0 + 0.566 370 614 759 21;
  • 3) 0.566 370 614 759 21 × 2 = 1 + 0.132 741 229 518 42;
  • 4) 0.132 741 229 518 42 × 2 = 0 + 0.265 482 459 036 84;
  • 5) 0.265 482 459 036 84 × 2 = 0 + 0.530 964 918 073 68;
  • 6) 0.530 964 918 073 68 × 2 = 1 + 0.061 929 836 147 36;
  • 7) 0.061 929 836 147 36 × 2 = 0 + 0.123 859 672 294 72;
  • 8) 0.123 859 672 294 72 × 2 = 0 + 0.247 719 344 589 44;
  • 9) 0.247 719 344 589 44 × 2 = 0 + 0.495 438 689 178 88;
  • 10) 0.495 438 689 178 88 × 2 = 0 + 0.990 877 378 357 76;
  • 11) 0.990 877 378 357 76 × 2 = 1 + 0.981 754 756 715 52;
  • 12) 0.981 754 756 715 52 × 2 = 1 + 0.963 509 513 431 04;
  • 13) 0.963 509 513 431 04 × 2 = 1 + 0.927 019 026 862 08;
  • 14) 0.927 019 026 862 08 × 2 = 1 + 0.854 038 053 724 16;
  • 15) 0.854 038 053 724 16 × 2 = 1 + 0.708 076 107 448 32;
  • 16) 0.708 076 107 448 32 × 2 = 1 + 0.416 152 214 896 64;
  • 17) 0.416 152 214 896 64 × 2 = 0 + 0.832 304 429 793 28;
  • 18) 0.832 304 429 793 28 × 2 = 1 + 0.664 608 859 586 56;
  • 19) 0.664 608 859 586 56 × 2 = 1 + 0.329 217 719 173 12;
  • 20) 0.329 217 719 173 12 × 2 = 0 + 0.658 435 438 346 24;
  • 21) 0.658 435 438 346 24 × 2 = 1 + 0.316 870 876 692 48;
  • 22) 0.316 870 876 692 48 × 2 = 0 + 0.633 741 753 384 96;
  • 23) 0.633 741 753 384 96 × 2 = 1 + 0.267 483 506 769 92;
  • 24) 0.267 483 506 769 92 × 2 = 0 + 0.534 967 013 539 84;
  • 25) 0.534 967 013 539 84 × 2 = 1 + 0.069 934 027 079 68;
  • 26) 0.069 934 027 079 68 × 2 = 0 + 0.139 868 054 159 36;
  • 27) 0.139 868 054 159 36 × 2 = 0 + 0.279 736 108 318 72;
  • 28) 0.279 736 108 318 72 × 2 = 0 + 0.559 472 216 637 44;
  • 29) 0.559 472 216 637 44 × 2 = 1 + 0.118 944 433 274 88;
  • 30) 0.118 944 433 274 88 × 2 = 0 + 0.237 888 866 549 76;
  • 31) 0.237 888 866 549 76 × 2 = 0 + 0.475 777 733 099 52;
  • 32) 0.475 777 733 099 52 × 2 = 0 + 0.951 555 466 199 04;
  • 33) 0.951 555 466 199 04 × 2 = 1 + 0.903 110 932 398 08;
  • 34) 0.903 110 932 398 08 × 2 = 1 + 0.806 221 864 796 16;
  • 35) 0.806 221 864 796 16 × 2 = 1 + 0.612 443 729 592 32;
  • 36) 0.612 443 729 592 32 × 2 = 1 + 0.224 887 459 184 64;
  • 37) 0.224 887 459 184 64 × 2 = 0 + 0.449 774 918 369 28;
  • 38) 0.449 774 918 369 28 × 2 = 0 + 0.899 549 836 738 56;
  • 39) 0.899 549 836 738 56 × 2 = 1 + 0.799 099 673 477 12;
  • 40) 0.799 099 673 477 12 × 2 = 1 + 0.598 199 346 954 24;
  • 41) 0.598 199 346 954 24 × 2 = 1 + 0.196 398 693 908 48;
  • 42) 0.196 398 693 908 48 × 2 = 0 + 0.392 797 387 816 96;
  • 43) 0.392 797 387 816 96 × 2 = 0 + 0.785 594 775 633 92;
  • 44) 0.785 594 775 633 92 × 2 = 1 + 0.571 189 551 267 84;
  • 45) 0.571 189 551 267 84 × 2 = 1 + 0.142 379 102 535 68;
  • 46) 0.142 379 102 535 68 × 2 = 0 + 0.284 758 205 071 36;
  • 47) 0.284 758 205 071 36 × 2 = 0 + 0.569 516 410 142 72;
  • 48) 0.569 516 410 142 72 × 2 = 1 + 0.139 032 820 285 44;
  • 49) 0.139 032 820 285 44 × 2 = 0 + 0.278 065 640 570 88;
  • 50) 0.278 065 640 570 88 × 2 = 0 + 0.556 131 281 141 76;
  • 51) 0.556 131 281 141 76 × 2 = 1 + 0.112 262 562 283 52;
  • 52) 0.112 262 562 283 52 × 2 = 0 + 0.224 525 124 567 04;
  • 53) 0.224 525 124 567 04 × 2 = 0 + 0.449 050 249 134 08;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.141 592 653 689 802 5(10) =

0.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 1001 0010 0(2)

5. Positive number before normalization:

3.141 592 653 689 802 5(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 1001 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.141 592 653 689 802 5(10) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 1001 0010 0(2) =

11.0010 0100 0011 1111 0110 1010 1000 1000 1111 0011 1001 1001 0010 0(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1100 1001 00(2) × 21


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1100 1001 00


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1024(10) =

100 0000 0000(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1100 1001 00 =

1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1100 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1100 1001


Decimal number 3.141 592 653 689 802 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1001 0010 0001 1111 1011 0101 0100 0100 0111 1001 1100 1100 1001

» Convert decimal 3.141 592 653 689 796 3 to 64 bit double precision IEEE 754 binary floating point representation standard

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100